The least integer whose multiplication with 588 leads to a perfect square is<\/p>\n
[amp_mcq option1=”2″ option2=”3″ option3=”4″ option4=”7″ correct=”option2″]<\/p>\n
To make 588 a perfect square by multiplication, we need to multiply it by an integer such that all the exponents in the resulting prime factorization are even. The least integer whose multiplication with 588 leads to a perfect square is [amp_mcq option1=”2″ option2=”3″ option3=”4″ option4=”7″ correct=”option2″] This question was previously asked in UPSC CAPF – 2013 Download PDFAttempt Online A perfect square is an integer that is the square of an integer. In terms of prime factorization, a number is a perfect … <\/p>\n
\nThe exponents in 588 are: 2 (for prime 2), 1 (for prime 3), and 2 (for prime 7).
\nThe exponents of 2 and 7 are already even.
\nThe exponent of 3 is 1, which is odd. To make it even, we need to multiply by 3 raised to an odd power. The least such power is 1. So, we must multiply by at least 3\u00b9.
\nMultiplying 588 by 3:
\n588 \u00d7 3 = (2\u00b2 \u00d7 3\u00b9 \u00d7 7\u00b2) \u00d7 3\u00b9 = 2\u00b2 \u00d7 3\u00b9\u207a\u00b9 \u00d7 7\u00b2 = 2\u00b2 \u00d7 3\u00b2 \u00d7 7\u00b2.
\nThe exponents are now 2, 2, and 2, which are all even.
\nThe resulting number is (2 \u00d7 3 \u00d7 7)\u00b2 = 42\u00b2, which is a perfect square.
\nThe least integer whose multiplication with 588 leads to a perfect square is 3.<\/section>\n